Coursera ML(5)-Logistic Regressi

作者: mmmwhy | 来源:发表于2017-04-15 16:50 被阅读64次

    [线性回归算法]iii.run,可用于房屋价格的估计及股票市场分析。 [Logistic Regression]iii.run(逻辑回归)是当前业界比较常用的机器学习方法,用于估计某种事物的可能性。比如某用户购买某商品的可能性,某病人患有某种疾病的可能性,以及某广告被用户点击的可能性等。相关公式推导在iii.run


    Stanford coursera Andrew Ng 机器学习课程编程作业(Exercise 2),作业下载链接貌似被墙了,下载链接放这。http://home.ustc.edu.cn/~mmmwhy/machine-learning-ex2.zip

    预备知识

    这里应该分为 正常、过拟合和欠拟合,三种情况。

    • Cost Function
      {% raw %} $$J(\theta) = - \frac{1}{m} \sum_{i=1}^m \large[ y^{(i)}\ \log (h_\theta (x^{(i)})) + (1 - y^{(i)})\ \log (1 - h_\theta(x^{(i)}))\large] + \frac{\lambda}{2m}\sum_{j=1}^n \theta_j^2$${% endraw %}

    • Gradient Descent
      {% raw %} $$\begin{align} & \text{Repeat}\ \lbrace \newline & \ \ \ \ \theta_0 := \theta_0 - \alpha\ \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y{(i)})x_0{(i)} \newline & \ \ \ \ \theta_j := \theta_j - \alpha\ \left[ \left( \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y{(i)})x_j{(i)} \right) + \frac{\lambda}{m}\theta_j \right] &\ \ \ \ \ \ \ \ \ \ j \in \lbrace 1,2...n\rbrace\newline & \rbrace \end{align}$${% endraw %}

    • Grad
      {% raw %} $$Grad = \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y{(i)})x_j{(i)}+ \frac{\lambda}{m}\theta_j$${% endraw %}

    后边有一个$\frac{\lambda}{2m}\sum_{j=1}^n \theta_j^2$和$\frac{\lambda}{m}\theta_j$小尾巴,作用就是进行 Regularization,防止拟合过度。

    Logistic Regression

    题目介绍

    • you will build a logistic regression model to predict whether a student gets admitted into a university.(根据各科目分数预测该学生是否能录取)
    • For each training example, you have the applicant's scores on two exams and the admissions decision.
    • Your task is to build a classication model that estimates an applicant's probability of admission based the scores from those two exams.
    dataset

    python code

    from numpy import *
    import matplotlib.pyplot as plt
    from scipy import optimize
    
    def init(path):
        X,y = load_dataset(path) # 调用底下那个东西
        m, n = shape(X)
        initial_theta = zeros(n + 1)
        return X,y,m,n,initial_theta
        
    def load_dataset(path):
        data = loadtxt(path, delimiter=',')
        X = data[:,:2]
        y = data[:, 2]
        return X,y
    
    def plotData(X,y):
        plt.plot(X[y==1][:,0],X[y==1][:,1],'k+',linewidth=2,)
        plt.plot(X[y==0][:,0],X[y==0][:,1],'ko',color='y',linewidth=2)
        plt.xlabel('科目一成绩', fontproperties='SimHei')
        plt.ylabel('科目二成绩', fontproperties='SimHei')
        plt.title('分数与录取的关系', fontproperties='SimHei')
        
        
    def sigmoid(X, theta):
        return 1 / (1 + exp(-dot(X, theta)))
    
    def get_cost(theta, X, y):
        J = sum((-y*log(sigmoid(X,theta)) - (1-y)*log(1-sigmoid(X,theta))))/len(X)
        return J
    
    def get_grad(theta, X, y):
        return (sigmoid(X,theta) - mat(y))*X*(1/m)
    
    def plotDecisionBoundary(theta, X, y):
        plotData(X[:, 1:3], y)
        if X.shape[1] <= 3:
            plot_x = r_[X[:,2].min()-2,  X[:,2].max()+2]
            plot_y = - (theta[1]*plot_x + theta[0])/theta[2]
            plt.plot(plot_x, plot_y)
            plt.axis([30,100,30,100])
            plt.legend(['Accepted', 'Not Accepted', 'Decision Boundary'])
            plt.show()
        else:
            pass
    
        
    def predict(theta, X):  
        prob = sigmoid([1,45,85] , result[0])
        return prob
        
    if __name__=="__main__":
        path = 'C:\\Users\\wing\\Desktop\\machine-learning-ex2\\ex2\\ex2data1.txt'
        X,y,m,n,initial_theta = init(path)
        X = column_stack((ones(m), X))
        cost = get_cost(initial_theta, X, y)
        grad = get_grad(initial_theta, X, y)
    
        # obtain the optimal theta
        result = optimize.fmin_tnc(func=get_cost, x0=initial_theta, fprime=get_grad, args=(X, y))  
        get_cost(result[0], X, y)  
        # result = (array([-25.16131863,   0.20623159,   0.20147149]), 36, 0)
        # get_cost(result[0], X, y)  = 0.20349770158947464
        plotDecisionBoundary(result[0], X, y)
        print('For a student with scores 45 and 85, we predict an admission ' \
             'probability of %f\n'%predict(result[0], X))
    
    # For a student with scores 45 and 85, we predict an admission probability of 0.776291
    

    运行结果


    最后进行了一个测试,如果一个学生两门考试成绩,一门45分,另外一门85分,那么他被录取的概率为77%。幸亏是在外国,在中国这分数,连大专都考不上。

    Logistic Regression and Regularization

    题目

    • Suppose you are the product manager of the factory and you have the test results for some microchips on two di�erent tests.
    • 对于一批产品,有两个检测环节,通过检测结果判断产品是否合格。比如,宜家会有三十年床垫保证,那么如果确保床垫合格(用30年),我们只能通过一些检测,来推测产品是否合格。

    python code

    from numpy import *
    import matplotlib.pyplot as plt
    from scipy import optimize
    
    def init(path):
        X,y = load_dataset(path)
        dataplot(X,y)
        X = map_feature(X[:,0], X[:,1])
        initial_theta = zeros(size(X[1]))
        lam = 1
        return X,y,initial_theta,lam
    
    def load_dataset(path):
        data = loadtxt(path, delimiter=',')
        X = data[:,:2]
        y = data[:, 2]
        return X,y
        
    def dataplot(X,y):
        plt.plot(X[y==1][:,0],X[y==1][:,1],'k+',linewidth=2)
        plt.plot(X[y==0][:,0],X[y==0][:,1],'ko',color='y',linewidth=2)
        plt.legend([ 'y = 1','y = 0'])
    
    def sigmoid(X, theta):
        return 1 / (1 + exp(-dot(X, theta)))
    
    
    def map_feature(x1, x2):
        #X1, X2, X1 ** 2, X2 ** 2, X1*X2, X1*X2 ** 2, etc...
    
        x1.shape = (x1.size, 1)
        x2.shape = (x2.size, 1)
        degree = 6
        out = ones(shape=(x1[:, 0].size, 1))
        m, n = out.shape
        for i in range(1, degree + 1):
            for j in range(i + 1):
                r = (x1 ** (i - j)) * (x2 ** j)
                out = append(out, r, axis=1)
        return out
    
    def get_cost(theta, X, y,lam):
        hx = sigmoid(X,theta)
        thetaR = theta[1:]
        J = sum((-y*log(hx) - (1-y)*log(1-hx)))/len(X) \
            + (lam / (2.0 * len(X))) * (thetaR.T.dot(thetaR))
        return J
    
    def get_grad(theta, X, y,lam):
        reg = (lam/len(y))*theta
        reg[0] = 0
        grad = X.T.dot(sigmoid(X,theta)-y)/len(y)+reg
        return grad
    
    def plotDecisionBoundary(theta,lam):
        u = linspace(-1, 1.5, 50)
        v = linspace(-1, 1.5, 50)
        z = zeros(shape=(len(u), len(v)))
        for i in range(len(u)):
            for j in range(len(v)):
                z[i, j] = (map_feature(array(u[i]), array(v[j])).dot(array(theta)))
        z = z.T
        plt.contour(u, v, z)
        plt.title('lambda = %f' % lam)
        plt.xlabel('Microchip Test 1')
        plt.ylabel('Microchip Test 2')
        plt.axis([-0.85,1.1,-0.85,1.1])
        plt.legend(['y = 1', 'y = 0', 'Decision boundary'])
        plt.show()
        
    if __name__=="__main__":
        path = 'C:\\Users\\wing\\Desktop\\machine-learning-ex2\\ex2\\ex2data2.txt'
        X,y,initial_theta,lam = init(path)
        result = optimize.fmin_tnc(func=get_cost, x0=initial_theta, fprime=get_grad, args=(X, y,lam))  
        plotDecisionBoundary(result[0],lam)
    

    运算结果

    • 过拟合
      lambda=0。不考虑{% raw %} $\frac{\lambda}{2m}\sum_{j=1}^n \theta_j^2$和$\frac{\lambda}{m}\theta_j${% endraw %} ,我们可以看到图像已经被拟合过度。这样的答案没有通用性


    • 欠拟合
      lambda=10,欠拟合会导致数据的很多细节被抛弃。


    • 拟合较好
      lambda=1,准确性到91%左右,这个准确率算低的了吧,还有很大上升空间。


    Summary

    熊辉上课的时候,说机器学习需要调参数,参数很不好调,需要使用者对数据有极高的敏感度。

    参数lambda就是这种感觉,感觉真的是乱调一通,然后就发现,诶哟,好像还不错。

    参考链接:
    scipy.optimize.minimize
    Logistic regression
    Machine Learning Exercises In Python, Part 3
    machine-learning-with-python-logistic


    以上

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